The energy released in the fusion of $2 \mathrm{~kg}$ of hydrogen deep in the sun is $E_H$ and the energy released in the fission of $2 \mathrm{~kg}$ of ${ }^{235} \mathrm{U}$ is $E_U$. The ratio $\frac{E_H}{E_U}$ is approximately: (Consider the fusion reaction as $$4_1^1H+2 \mathrm{e}^{-} \rightarrow{ }_2^4 \mathrm{He}+2 v+6 \gamma+26.7 \mathrm{~MeV}$$, energy released in the fission reaction of ${ }^{235} \mathrm{U}$ is $200 \mathrm{~MeV}$ per fission nucleus and $\mathrm{N}_{\mathrm{A}}= 6.023 \times 10^{23})$

<p>To determine the ratio $\frac{E_H}{E_U}$, let's first consider the energy released in the fusion of hydrogen and the energy released in the fission of $^{235}U$.</p> <p>For the fusion reaction: </p> <p>The given reaction is:</p> <p>$$4_1^1H + 2 \mathrm{e}^{-} \rightarrow {}_2^4 \mathrm{He} + 2 \mathrm{\nu} + 6 \gamma + 26.7 \, \mathrm{MeV}$$</p> <p>This reaction shows that 4 hydrogen atoms and 2 electrons produce 1 helium atom, 2 neutrinos, and 6 gamma photons while releasing 26.7 MeV of energy.</p> <p>The mass of 1 mole of $H_1^1$ is approximately 1 gram, so 2 kg of hydrogen is equal to:</p> <p>$2 \times 10^3 \, \mathrm{g}$</p> <p>The number of moles of hydrogen in 2 kg is:</p> <p>$$\mathrm{N}_\text{moles} = \frac{2 \times 10^3}{1} = 2 \times 10^3 \, \mathrm{moles}$$</p> <p>The number of hydrogen atoms in 2 kg is:</p> <p>$$\mathrm{N}_\text{atoms} = N_A \times 2 \times 10^3 = 6.023 \times 10^{23} \times 2 \times 10^3 = 1.2046 \times 10^{27} \, \mathrm{atoms}$$</p> <p>Since 4 hydrogen atoms release 26.7 MeV, the total energy released ($E_H$) is:</p> <p>$E_H = \frac{1.2046 \times 10^{27}}{4} \times 26.7 \, \mathrm{MeV}$</p> <p>$E_H = 3.0115 \times 10^{26} \times 26.7 \, \mathrm{MeV}$</p> <p>$E_H \approx 8.04 \times 10^{27} \, \mathrm{MeV}$</p> <p>For the fission reaction: </p> <p>The energy released per fission of one $^{235}U$ nucleus is 200 MeV.</p> <p>The mass of 1 mole of $^{235}U$ is approximately 235 grams, so 2 kg of ${}^{235}U$ is equal to:</p> <p>$2 \times 10^3 \, \mathrm{g}$</p> <p>The number of moles of $^{235}U$ in 2 kg is:</p> <p>$$\mathrm{N}_\text{moles} = \frac{2 \times 10^3}{235} \approx 8.51 \, \mathrm{moles}$$</p> <p>The number of $^{235}U$ nuclei is:</p> <p>$$\mathrm{N}_\text{nuclei} = N_A \times 8.51 \approx 6.023 \times 10^{23} \times 8.51 \approx 5.12 \times 10^{24} \, \mathrm{nuclei}$$</p> <p>The total energy released ($E_U$) is:</p> <p>$$E_U = 200 \, \mathrm{MeV} \times 5.12 \times 10^{24} \approx 1.024 \times 10^{27} \, \mathrm{MeV}$$</p> <p>Finally, the ratio $\frac{E_H}{E_U}$ is:</p> <p>$$\frac{E_H}{E_U} \approx \frac{8.04 \times 10^{27}}{1.024 \times 10^{27}} \approx 7.85$$</p> <p>Therefore, the ratio is closest to option A: 7.62.</p>